Need help analyzing a circuit

The main simplifying assumption to make is that the opamp will use its output to keep its two inputs at the same voltage. Since the noninverting input is at 0 volts, the output will use the output darlington pair of PNP transistors to hold its - (inverting ) input at zero volts, also.

The second simplifying assumption is that the two inputs of the opamp draw no current.

So any current that arrives at from the positive control voltage (+4) through the 56k resistor must be sucked up by the 1k emitter resistor in such a way that the node between those two resistors stays at zero volts. The two transistors are connected to multiply the current gains of the transistors, so that the base current of Q2 is so small that you can neglect it. So essentially all the current through the

A capacitor is a device that allows charge to 'pile up' (technical term).

A pile of charge is a voltage.

The formal relation is

charge = capacitance times voltage

This is another way of saying

current = capacitance times 'rate of change of voltage'.

For the above, you know that the opamp will try to keep the junction between the 56k and 1k resistor at 0V, so the current through the 56k resistor will be

The capacitor is 10nF, so the 'rate of change' of voltage is

If the positive side of the capacitor starts at -12V, then to get to, say, -2V will take

The waveform is constantly increasing (it's linear), so the output, taken from the positive side of the capacitor, is a 'ramp', which must be brought back to -12V periorically by shorting across the capacitor, probably using a transistor.

Thanks for the explanation, but I have one question. I'm familiar with OPAMP's used with negative feedback and it sounds like this is what you are describing when you say that the OPAMP will attempt to keep the junction of th 56k and 1k at 0v. However, what confuses me is that the output of the OPAMP goes into two PNP transistors where the emitters are connected to the 1k resistor. In conventional electron flow, wouldn't the current be coming through the 1k, and through the emitters exiting through the base at the 47 ohm resistor and out through the collector connected to the 100 ohm resistor? In this arrangement I don't see how there can be negative feedback into the inverting pin since it would seem that the current couldn't flow through the base/emitter junctions back towards the

I must be misunderstanding something.

Thanks again.

When the opamp output voltage is two diode drops more negative than the right end of the 1k resistor, the Q2 begins to conduct, supplying base current to Q1 which also begins to conduct, and all that current passes through the 1k resistor, making its right end more negative than its right. This process continues till the voltage at the junction of the two resistors balances at zero volts, so that all the current passing through the 56k resistor also passes through the 1k resistor, while the voltage across the 56k resistor is 4 volts.

At the same time almost all that current also passes through the 100 ohm resistor. The only part missing is the small base current being injected by the opamp output into Q2 which becomes part of the emitter current of Q1 but does not make it to the collectors.

Ok, that makes sense. Thanks for the explanation.

Funny you should mention building the circuit to just play around with as I was planning to do just that! In fact, I just picked up a used scope that I'm hoping to use for this sort of thing. I'll also look into the simulators. I had downloaded an evaluation copy of one and played around with it, but I like the idea of working with the real circuit.

Ok, I've got some really dumb followup questions. If the OPAMP is going to keep the junction between the 56k and 1k resistor at 0V, what is the purpose of the 47 ohm resistor and the Darlington pair? If the current through the 56k resistor basically is used to charge the capacitor, what role do the other components play? Why the 1k resistor (why that particular size)? Why the 100 ohm resistor?

Thanks.

I'm not sure I would have included those components. The 47 ohm resistor looks like a current limit resistor that comes into play if the collectors of the transistors have no current path (like if the capacitor is allowed to charge up till the first collector to base junction swings to forward bias). The 100 ohm resistor is more problematic, unless the discharge switch you mentioned earlier connects between it and the collectors, so that the peak discharge current is limited by the 100 ohms. You have to think about all the strange and unusual operating conditions the circuit might experience to guess what the designer had in mind for some components.

One more thing. One nice way to get to play around with a circuit like this is to either build it and see what happens (not really possible unless you have a scope), or simulate it. Terry Pinnell has a great page on simulators (free and not free) that you can use. It's here

I like circuitmaker student for simple stuff, and pspice student for temperature dependent analysis. There is a also a free one which people here tend to post with, called LTSpice, at the linear site.

In the full circuit, there is a connection between the .01uf cap and the

In that case, I see no point to having the 100 ohm resistor. The transistors represent a limited source of current, whether charging the cap or dumping into the discharge pin of a 555.

The problem with simulators is they tend to lie if you screw up the parameters, or if the circuit contains things that are difficult to model, or for any number of reasons. Also, you often get into situations where the mathematical models they use don't converge, and you end up trying to figure out ways to fix the circuit just to get it to converge.

However, they let you 'see' things that are fairly difficult to see with a scope, like currents and power through circuit items. You can also show any number of varying voltages at once, thus overcoming the two-trace limitations of many oscilloscopes.

The darlington minimizes current flow into the opamp for control. With a normal PNP, you have a 1% error. With a darlington, you'll have a 0.01% error.

You could minimize error even more using a p-channel JFET, which has essentially no leakage when the cap is charging, but also allows the current to be diverted when the cap reaches 0V. In order for this to matter, you need to use a good capacitor, such as a teflon cap.

That's interesting, could you explain that a little more? Is this something that is talked about in the data sheets for the 734 OPAMP? I read something somewhere last night about this op amp having problems at very low voltage levels. Does this configuration help the op amp produce accurate lower voltages. Is that what you meant by the 0.01% error? Also, would this configuration help with making the current converter more resistant to temperature issues? This particular circuit is part of an analog music synthesizer so they would want a particular voltage to produce a particular frequency with consistency.

Thanks

No, the issue I was referring to is current leakage for control. The way a bipolar transistor works is that some of the current is taken by the diode formed by the emitter-base (for PNP). That current flow allows a much larger current to flow from emitter to collector. For a normal transistor, the ratio of collector current to base current is called 'beta'. It's usually between 15 and 300.

Thus, if you are depending on the fact that the 56k resistor is delivering exactly 1/14000 A, some of that current will be diverted to the base rather than delivered to the capacitor. A ballpark figure is

For a darlington pair, however, the control current for the 'first' transistor is again diverted to the output, using 1/100 of *that* current. Thus, the real control current which flows into the opamp is